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# The addition problem above shows four of the 24 different in

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The addition problem above shows four of the 24 different in [#permalink]

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03 Nov 2010, 00:34
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1,234
1,243
1,324
.....
....
+4,321

The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers?

A. 24,000
B. 26,664
C. 40,440
D. 60,000
E. 66,660
[Reveal] Spoiler: OA

Last edited by Bunuel on 06 Nov 2012, 03:17, edited 1 time in total.
Renamed the topic and edited the question.

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03 Nov 2010, 00:45
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student26 wrote:
1,234
1,243
1,324
.....
....
+4,321

The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers?

A.24,000
B.26,664
C.40,440
D.60,000
E.66,660

Using the symmetry in the numbers involved (All formed using all possible combinations of 1,2,3,4), and we know there are 24 of them. We know there will be 6 each with the units digits as 1, as 2, as 3 and as 4. And the same holds true of the tens, hundreds and thousands digit.

The sum is therefore = (1 + 10 + 100 + 1000) * (1*6 +2*6 +3*6 +4*6) = 1111 * 6 * 10 = 66660

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08 Feb 2011, 06:26
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1,2,3,4 can be arranged in 4! = 24 ways

The units place of all the integers will have six 1's, six 2's, six 3's and six 4's
Likewise,
The tens place of all the integers will have six 1's, six 2's, six 3's and six 4's
The hundreds place of all the integers will have six 1's, six 2's, six 3's and six 4's
The thousands place of all the integers will have six 1's, six 2's, six 3's and six 4's

Addition always start from right(UNITS) to left(THOUSANDS);

Units place addition; 6(1+2+3+4) = 60.
Unit place of the result: 0
carried over to tens place: 6

Tens place addition; 6(1+2+3+4) = 60 + 6(Carried over from Units place) = 66
Tens place of the result: 6
carried over to hunderes place: 6

Hundreds place addition; 6(1+2+3+4) = 60 + 6(Carried over from tens place) = 66
Hundreds place of the result: 6
carried over to thousands place: 6

Thousands place addition; 6(1+2+3+4) = 60 + 6(Carried over from hundreds place) = 66
Thousands place of the result: 6
carried over to ten thousands place: 6

Ten thousands place of the result: 0+6(Carried over from thousands place) = 6

Result: 66660

Ans: "E"
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08 Feb 2011, 06:48
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Merging similar topics.

Formulas for such kind of problems (just in case):

1. Sum of all the numbers which can be formed by using the $$n$$ digits without repetition is: $$(n-1)!*(sum \ of \ the \ digits)*(111... \ n \ times)$$.

2. Sum of all the numbers which can be formed by using the $$n$$ digits (repetition being allowed) is: $$n^{n-1}*(sum \ of \ the \ digits)*(111... \ n \ times)$$.

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22 Jan 2013, 10:44
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Bunuel wrote:
Merging similar topics.

Formulas for such kind of problems (just in case):

1. Sum of all the numbers which can be formed by using the $$n$$ digits without repetition is: $$(n-1)!*(sum \ of \ the \ digits)*(111... \ n \ times)$$.

2. Sum of all the numbers which can be formed by using the $$n$$ digits (repetition being allowed) is: $$n^{n-1}*(sum \ of \ the \ digits)*(111... \ n \ times)$$.

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sum-of-3-digit-s-78143.html

Could you tell me the way to calculate the sum which the repetition is allowed? For example: from 1,2,3,4. how can we calculate the sum of four digit number that formed from 1,2,3,4 and repetition is allowed?
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22 Jan 2013, 21:07
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hellscream wrote:

Could you tell me the way to calculate the sum which the repetition is allowed? For example: from 1,2,3,4. how can we calculate the sum of four digit number that formed from 1,2,3,4 and repetition is allowed?

The logic is no different from 'no repetition allowed' question. The only thing different is the number of numbers you can make.

How many numbers can you make using the four digits 1, 2, 3 and 4 if repetition is allowed?
You can make 4*4*4*4 = 256 numbers (there are 4 options for each digit)

1111
1112
1121
... and so on till 4444

By symmetry, each digit will appear equally in each place i.e. in unit's place, of the 256 numbers, 64 will have 1, 64 will have 2, 64 will have 3 and 64 will have 4.
Same for 10s, 100s and 1000s place.

Sum = 1000*(64*1 + 64*2 + 64*3 + 64*4) + 100*(64*1 + 64*2 + 64*3 + 64*4) + 10*(64*1 + 64*2 + 64*3 + 64*4) + 1*(64*1 + 64*2 + 64*3 + 64*4)
= (1000 + 100 + 10 + 1)(64*1 + 64*2 + 64*3 + 64*4)
= 1111*64*10 = 711040

or use the formula given by Bunuel above:
Sum of all the numbers which can be formed by using the digits (repetition being allowed) is:$$n^{n-1}$$*Sum of digits*(111...n times)
=$$4^3*(1+2+3+4)*(1111) = 711040$$ (Same calculation as above)
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17324 [2], given: 232 Intern Joined: 21 May 2013 Posts: 1 Kudos [?]: [0], given: 0 The addition problem above shows four of the 24 different intege [#permalink] ### Show Tags 21 May 2013, 07:35 This can be solved much easier by realizing that, since the number of four term permutations is 4!, and that summing the a sequence to its reverse gives 1234 +4321 = 5555 1243 +3421 = 5555 we may see that there are 4!/2 pairings we can make, giving us 5555(12) = 66660 Kudos [?]: [0], given: 0 Intern Joined: 25 Jul 2014 Posts: 18 Kudos [?]: 37 [2], given: 52 Concentration: Finance, General Management GPA: 3.54 WE: Asset Management (Venture Capital) Re: The addition problem above shows four of the 24 different in [#permalink] ### Show Tags 28 Sep 2014, 10:41 2 This post received KUDOS For those who could not memorize the formular, you can guess the answer in 30 secs: Since we have 24 numbers, we will have 6 of 1 thousand something, 6 of 2 thousand something, 6 of 3 thousand something, and 6 of 4 thousand something So, 6x1(thousand something) = 6 (thousand something) 6x2(thousand something) = 12 (thousand something) 6x3(thousand something) = 18 (thousand something) 6x4(thousand something) = 24 (thousand something) Add them all 6+12 +18 + 24 = 60 (thousand something) ----> E Kudos [?]: 37 [2], given: 52 Intern Joined: 24 Apr 2016 Posts: 6 Kudos [?]: 1 [0], given: 895 Re: The addition problem above shows four of the 24 different in [#permalink] ### Show Tags 15 Aug 2016, 14:38 each term is repeating 6 times..(total 24/4=4) now at unit...each term will repeat 6 times... (1x6+ 2x6 + 3x6 + 4x6 = 60) , so unit digit is "0" and 6 remaining. repeating same.....total of 24 ten digit will be 60 + 6 from total of unit ,so ten digit will be 6. only E has 60 as last two digits. Kudos [?]: 1 [0], given: 895 VP Status: Learning Joined: 20 Dec 2015 Posts: 1069 Kudos [?]: 69 [1], given: 532 Location: India Concentration: Operations, Marketing GMAT 1: 670 Q48 V36 GRE 1: 314 Q157 V157 GPA: 3.4 WE: Manufacturing and Production (Manufacturing) Re: The addition problem above shows four of the 24 different in [#permalink] ### Show Tags 14 Sep 2017, 07:19 1 This post received KUDOS student26 wrote: 1,234 1,243 1,324 ..... .... +4,321 The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers? A. 24,000 B. 26,664 C. 40,440 D. 60,000 E. 66,660 Each digit will come at the respective place i.e units,tens, hundreds , thousands So calculate sum of each digit for the all the places for 4=4000+400+40+4 3=3000+300+30+3 2=2000+200+20+2 1=1000+100+10+1 Now calculate the sum of the sums of these digits =11110 Now we know that each digit is used 6 times therefore we have to multiply with 6 6*11110=66660 Hence E is our answer . _________________ We are more often frightened than hurt; and we suffer more from imagination than from reality Kudos [?]: 69 [1], given: 532 Intern Joined: 16 Jul 2011 Posts: 39 Kudos [?]: 1 [0], given: 163 Concentration: Marketing, Real Estate GMAT 1: 550 Q37 V28 GMAT 2: 610 Q43 V31 Re: The addition problem above shows four of the 24 different in [#permalink] ### Show Tags 02 Oct 2017, 14:41 VeritasPrepKarishma wrote: hellscream wrote: Could you tell me the way to calculate the sum which the repetition is allowed? For example: from 1,2,3,4. how can we calculate the sum of four digit number that formed from 1,2,3,4 and repetition is allowed? The logic is no different from 'no repetition allowed' question. The only thing different is the number of numbers you can make. How many numbers can you make using the four digits 1, 2, 3 and 4 if repetition is allowed? You can make 4*4*4*4 = 256 numbers (there are 4 options for each digit) 1111 1112 1121 ... and so on till 4444 By symmetry, each digit will appear equally in each place i.e. in unit's place, of the 256 numbers, 64 will have 1, 64 will have 2, 64 will have 3 and 64 will have 4. Same for 10s, 100s and 1000s place. Sum = 1000*(64*1 + 64*2 + 64*3 + 64*4) + 100*(64*1 + 64*2 + 64*3 + 64*4) + 10*(64*1 + 64*2 + 64*3 + 64*4) + 1*(64*1 + 64*2 + 64*3 + 64*4) = (1000 + 100 + 10 + 1)(64*1 + 64*2 + 64*3 + 64*4) = 1111*64*10 = 711040 or use the formula given by Bunuel above: Sum of all the numbers which can be formed by using the digits (repetition being allowed) is:$$n^{n-1}$$*Sum of digits*(111...n times) =$$4^3*(1+2+3+4)*(1111) = 711040$$ (Same calculation as above) How about using the Average*no. of terms formula where the average is found by adding the smallest number and the largest number divided by 2. It worked for this question. My question is will it work for all such types of questions? _________________ "The fool didn't know it was impossible, so he did it." Kudos [?]: 1 [0], given: 163 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7668 Kudos [?]: 17324 [1], given: 232 Location: Pune, India Re: The addition problem above shows four of the 24 different in [#permalink] ### Show Tags 02 Oct 2017, 20:57 1 This post received KUDOS Expert's post sam2016 wrote: VeritasPrepKarishma wrote: hellscream wrote: Could you tell me the way to calculate the sum which the repetition is allowed? For example: from 1,2,3,4. how can we calculate the sum of four digit number that formed from 1,2,3,4 and repetition is allowed? The logic is no different from 'no repetition allowed' question. The only thing different is the number of numbers you can make. How many numbers can you make using the four digits 1, 2, 3 and 4 if repetition is allowed? You can make 4*4*4*4 = 256 numbers (there are 4 options for each digit) 1111 1112 1121 ... and so on till 4444 By symmetry, each digit will appear equally in each place i.e. in unit's place, of the 256 numbers, 64 will have 1, 64 will have 2, 64 will have 3 and 64 will have 4. Same for 10s, 100s and 1000s place. Sum = 1000*(64*1 + 64*2 + 64*3 + 64*4) + 100*(64*1 + 64*2 + 64*3 + 64*4) + 10*(64*1 + 64*2 + 64*3 + 64*4) + 1*(64*1 + 64*2 + 64*3 + 64*4) = (1000 + 100 + 10 + 1)(64*1 + 64*2 + 64*3 + 64*4) = 1111*64*10 = 711040 or use the formula given by Bunuel above: Sum of all the numbers which can be formed by using the digits (repetition being allowed) is:$$n^{n-1}$$*Sum of digits*(111...n times) =$$4^3*(1+2+3+4)*(1111) = 711040$$ (Same calculation as above) How about using the Average*no. of terms formula where the average is found by adding the smallest number and the largest number divided by 2. It worked for this question. My question is will it work for all such types of questions? Sum of terms = Average * Number of terms holds for all sets of numbers Average = (First term + Last term)/2 holds for an Arithmetic Progression only _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: The addition problem above shows four of the 24 different in [#permalink]

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03 Oct 2017, 12:30
VeritasPrepKarishma wrote:
sam2016 wrote:

How about using the Average*no. of terms formula where the average is found by adding the smallest number and the largest number divided by 2. It worked for this question. My question is will it work for all such types of questions?

Sum of terms = Average * Number of terms
holds for all sets of numbers

Average = (First term + Last term)/2
holds for an Arithmetic Progression only

Yes, you are right. But what was the arithmetic progression in this question?
And do such types of questions always have arithmetic progression so that I can use the above-mentioned formula for such questions?
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Re: The addition problem above shows four of the 24 different in [#permalink]

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03 Oct 2017, 21:45
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sam2016 wrote:
VeritasPrepKarishma wrote:
sam2016 wrote:

How about using the Average*no. of terms formula where the average is found by adding the smallest number and the largest number divided by 2. It worked for this question. My question is will it work for all such types of questions?

Sum of terms = Average * Number of terms
holds for all sets of numbers

Average = (First term + Last term)/2
holds for an Arithmetic Progression only

Yes, you are right. But what was the arithmetic progression in this question?
And do such types of questions always have arithmetic progression so that I can use the above-mentioned formula for such questions?

There isn't and hence, I haven't used this formula. Note that Avg concept will not work when the digits are say 1, 2, 4, 6.
It works in this case because of the symmetry of the digits 1, 2, 3 and 4.
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Re: The addition problem above shows four of the 24 different in   [#permalink] 03 Oct 2017, 21:45
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